to find the height of a building, a 7 -foot basketball player places a mirror on the ground 100 ft from the base of a building. standing 8 ft on the other side of the mirror, he looks into the mirror and sees the top of the building. how tall is the building

Answer:

B) Decay

Step-by-step explanation:

This is a decay function because if you look at the number being raised to the power of t it is (1-0.43) meaning that the factor is less than 1, and numbers less than 1 result in a decay function. Whereas function that have a rate of 1 of more will be growth.

Answer:

Step-by-step explanation:

Answer:

(a) What percentage of women would not be able to use the original ejection seats designed for men?

To answer this question, we need to find the percentage of women whose weight is outside the range of 140 lb to 201 lb, which is the weight range for which the original ejection seats were designed. We can do this by standardizing the weight distribution of women using the formula:

z = (x - μ) / σ

where z is the standard score, x is the weight of a woman, μ is the mean weight of women, and σ is the standard deviation of the weight of women.

Substituting the given values, we get:

z = (140 - 162) / 46 = -0.478

z = (201 - 162) / 46 = 0.848

Using a standard normal table or a calculator, we can find that the percentage of women whose weight is below 140 lb (corresponding to a z-score of -0.478) is 31.45%. Similarly, the percentage of women whose weight is above 201 lb (corresponding to a z-score of 0.848) is 19.34%. Therefore, the percentage of women who would not be able to use the original ejection seats designed for men is:

31.45% + 19.34% = 50.79%

(b) What weight range would accommodate 99.7% of the female pilots?

To find the weight range that would accommodate 99.7% of the female pilots, we need to find the z-scores that correspond to the 0.15% and 99.85% percentiles of the standard normal distribution. Using a standard normal table or a calculator, we can find that these z-scores are approximately -2.97 and 2.97, respectively.

Substituting these z-scores into the standardization formula, we get:

-2.97 = (x - 162) / 46

x = 63.88 lb

and

2.97 = (x - 162) / 46

x = 260.12 lb

Therefore, the weight range that would accommodate 99.7% of the female pilots is approximately 63.88 lb to 260.12 lb.

(c) What percentage of women would need ejection seats redesigned for them?

To find the percentage of women who would need ejection seats redesigned for them, we need to find the percentage of women whose weight is outside the weight range that would accommodate 99.7% of the female pilots. Using the values from part (b), we can see that the weight range that would accommodate 99.7% of the female pilots is approximately 63.88 lb to 260.12 lb. Therefore, the percentage of women who would need ejection seats redesigned for them is:

100% - 99.7% = 0.3%

The correct representations of the inequality –3(2x – 5) < 5(2 – x) are:

-6x + 15 < 10 – 5x and x < 5

What is the inequality equation?

An inequality equation is a mathematical statement that compares two expressions using an inequality symbol such as < (less than), > (greater than), ≤ (less than or equal to), or ≥ (greater than or equal to).

Option 1: x < 5 – This is one of the correct representations. It is obtained by solving the inequality as follows:

-3(2x – 5) < 5(2 – x)

-6x + 15 < 10 – 5x (Distribute on the left and right sides)

-x < -5

x > 5 (Multiply both sides by -1 and reverse the inequality)

Option 2: -6x – 5 < 10 – x – This is not a correct representation. It is obtained by distributing on the left side but incorrectly distributing on the right side.

Option 3: -6x + 15 < 10 – 5x – This is one of the correct representations. It is obtained by distributing on the left and right sides, and then simplifying.

Option 4: x + 6 < 5x – 10 – This is not a correct representation. It is obtained by incorrectly distributing on the right side.

As for the number line representations:

The number line from negative 3 to 3 in increments of 1 with an open circle at 5 and a bold line starting at 5 and pointing to the right represents the solution x < 5, which is correct.

The number line from negative 3 to 3 in increments of 1 with an open circle at negative 5 and a bold line starting at negative 5 and pointing to the left does not represent the correct solution. There is no negative solution to this inequality, so there should not be an open circle or a bold line pointing to the left.

Hence, The correct representations of the inequality –3(2x – 5) < 5(2 – x) are:

-6x + 15 < 10 – 5x and x < 5

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The sampling technique used in this scenario is stratified random sampling.

This is because Kane grouped the population of licensed nurses in Florida into two distinct strata, namely urban and nonurban areas, based on a specific characteristic (geographical location). Then, he randomly selected 250 nurses from each stratum, resulting in a total sample size of 500 nurses.

The purpose of stratified random sampling is to ensure that the sample is representative of the population by including participants from each stratum in proportion to their representation in the population. This allows for more accurate and precise estimates of population parameters than simple random sampling, especially when there are significant differences between strata.

By stratifying the population and randomly selecting participants from each stratum, Kane can draw more reliable conclusions about the attitudes of licensed nurses toward evidence-based practice in urban and nonurban areas of Florida.

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The value of x is 40 given that the two prisms are similar.

If the two prisms are comparable, then their corresponding sides will be proportionately the same.

x / 20 =  10 / 5

x/ 20 = 2

x = 2 * 20

x = 40

Therefore, the value of x is 40 given that the two prisms are similar.

A prism is a geometric shape that has two parallel and congruent bases that are connected by a set of rectangular or parallelogram faces. When light enters a prism, it is refracted, or bent, due to the different angles of the surfaces of the prism.

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the question lack some information.

See more detail of the question on the attached image.

The value of the expression m + (7.9) is 10.4. when m= 2.5 and n = 5

from the question, we have the following parameters that can be used in our computation:

m + (7.9)

The expression m + (7.9) means that we need to add the value of m to 7.9.

Substituting m = 2.5, we get:

m + (7.9) = 2.5 + 7.9 = 10.4

Therefore, when m = 2.5, the value of the expression m + (7.9) is 10.4.

Note that the value of n is not used in this expression since it is not included in the expression to be evaluated.

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There are 90 positive odd integers less than 1000 that can be formed from the digits 0, 1, 2, 3, 4, 5, and 6 by choosing any of the three odd digits as the last digit, and any of the remaining digits for the first and second positions.

To form a positive odd integer, the last digit must be an odd number, i.e., 1, 3, 5. For each of these three possible last digits, we can choose any of the remaining six digits for the first position, and any of the remaining five digits for the second position (since we cannot repeat digits in forming a positive integer).

Therefore, the total number of positive odd integers less than 1000 that can be formed from the digits 0, 1, 2, 3, 4, 5, and 6 is:

3 × 6 × 5 = 90

So there are 90 positive odd integers less than 1000 that can be formed from the given digits.

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The probability that at least one of the next ten earthquakes to strike this region will exceed 5.0 on the Richter scale is approximately 0.867.

To solve this problem, we need to use the cumulative distribution function (CDF) of the exponential distribution.

Let X be the magnitude of earthquakes in this region, and [tex]X ~ exp(1/2.4).[/tex]

To find the probability that an earthquake striking this region will exceed 3.0 on the Richter scale, we need to calculate P(X > 3).

[tex]P(X > 3) = 1 - P(X ≤ 3)\\= 1 - F(3) (where F is the CDF of X)\\= 1 - (1 - e^(-3/2.4))\\= e^(-1.25)[/tex]

≈ 0.286

Therefore, the probability that an earthquake striking this region will exceed 3.0 on the Richter scale is approximately 0.286.

To find the probability that an earthquake striking this region will fall between 2.0 and 3.0 on the Richter scale, we need to calculate P(2 < X ≤ 3).

[tex]P(2 < X ≤ 3) = F(3) - F(2)= (1 - e^(-3/2.4)) - (1 - e^(-2/2.4))[/tex]

≈ 0.285

Therefore, the probability that an earthquake struck this region will fall between 2.0 and 3.0 on the Richter scale is approximately 0.285.

To find the probability that at least one of the next ten earthquakes to strike this region will exceed 5.0 on the Richter scale, we need to use the complement rule and the fact that the number of earthquakes is a Poisson distribution with mean 10/2.4 = 4.17. Let Y be the number of earthquakes that exceed 5.0 on the Richter scale.

[tex]P(Y ≥ 1) = 1 - P(Y = 0)\\= 1 - e^(-4.17) * (4.17^0 / 0!)[/tex]

≈ 0.867

Therefore, the probability that at least one of the next ten earthquakes to strike this region will exceed 5.0 on the Richter scale is approximately 0.867.

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Ms. Wilson will receive a amount given by option D. She will receive a refund of $258.

Total taxable income of Ms. Wilson's = $36,201

Status filled by Ms. Wilson is Single.

Amount already paid as federal tax = $5344

From the table,

Taxable amount to be paid by single .

Range of income is 36,200 to 36250 is equal to $5086

Here federal tax paid by Ms. Wilson is higher then tax to be paid.

This implies,

She will receive a refund.

Refund amount = Federal tax amount - Tax to be paid

Substitute the value we have,

⇒Refund amount = $5344 - $5086

⇒Refund amount =$258

Therefore, the Ms. Wilson will receive her refund represented by option D. She will receive a refund of $258.

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Answer:

x = 4

Step-by-step explanation:

Answer:

6x^(3/2)

Step-by-step explanation:

To simplify the expression (2√x) × (3³√x), we can first use the properties of exponents to rewrite the cube root as a fractional exponent:

3³√x = x^(1/3)^3 = x^(1)

Using this, we can rewrite the expression as:

(2√x) × (3³√x) = 2x^(1/2) × 3x^(1)

To multiply these two terms, we can add the exponents since the bases are the same:

2x^(1/2) × 3x^(1) = 6x^(1/2 + 1) = 6x^(3/2)

Therefore, the simplified expression is 6x^(3/2).

This equation is clearly not true, which means there is no real value of x that satisfies it. Therefore, the system of equations represented by line A and line B has no real solutions, the correct option is C.

Line A has a slope of 7, which means any line parallel to it will also have a slope of 7. Therefore, line B can be represented by the equation y = 7x + b, where b is the y-intercept that we need to find.

We know that line B passes through the point (2, 24), so we can substitute these values into the equation to get:

24 = 7(2) + b

Simplifying the right side gives:

24 = 14 + b

Subtracting 14 from both sides gives:

b = 10

Therefore, the equation of line B is y = 7x + 10.

Now we need to find the point of intersection between line A and line B, which is the solution to the system of equations represented by these two lines. To accomplish this, we can put the two equations equal to one another and find x:

7x + 12 = 7x + 10

Subtracting 7x from both sides gives:

12 = 10

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They can get 32 portions of 1/8 lb each from a 4-pound block of Parmesan cheese.

To determine how many portions can be made from a 4-pound block of Parmesan cheese at the market, follow these steps:
1. Identify the size of each portion: 1/8 lb
2. Identify the total weight of the cheese block: 4 lb
3. Divide the total weight by the portion size to find the number of portions:

4 lb ÷ 1/8 lb

To do this division, you can multiply 4 by the reciprocal of 1/8 (which is 8/1):
4 × 8/1 = 32.

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A linear function has one independent variable and one dependent variable..

therefore, all are linear except y=x²-6

In order to determine the chi-squared value and the predicted level of significance as represented by the test, we need to conduct a hypothesis test. We will use the chi-squared test for a population variance. The null hypothesis is that the population variance is equal to 20 minutes squared (the square of the standard deviation), and the alternative hypothesis is that the population variance is greater than 20 minutes squared.

Therefore, the chi-squared value is calculated as:χ2 = (n - 1) × s2 / σ20where n is the sample size, s is the sample standard deviation, and σ0 is the hypothesized population standard deviation. Substituting the values given in the question, we get:χ2 = (7 - 1) × 29² / 20²= 21.19.

The degrees of freedom for this test are df = n - 1 = 6.Using a chi-squared distribution table with 6 degrees of freedom, we can find the predicted level of significance. For a chi-squared value of 21.19 and 6 degrees of freedom, the level of significance is less than 0.005. Therefore, we reject the null hypothesis and conclude that there is sufficient evidence to support the claim that the population variance is greater than 20 minutes squared.

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The area of the regular polygon is  [tex]49\sqrt3[/tex] square unit.

What is area?

The region that an object's shape defines as its area. The area of a figure or any other two-dimensional geometric shape in a plane is how much space it occupies.

Here the given triangle ,

Apothem a = [tex]\frac{7\sqrt3}{3}[/tex]

Now side length s = [tex]2\sqrt3 a[/tex]

=> side length s = [tex]2\sqrt3\times \frac{7\sqrt3}{3} = \frac{14\times3}{3}[/tex] = 14

Now area of equilateral triangle A = [tex]\frac{\sqrt3 s^2}{4}[/tex] square unit.

=> A = [tex]\frac{\sqrt3\times14^2}{4}[/tex] = [tex]49\sqrt3[/tex] square unit.

Hence the area of the regular polygon is  [tex]49\sqrt3[/tex] square unit.

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Eden consumed a total of 9.75 quarts of sports drinks during both practices.

Eden brings a 0.75-quart bottle of sports drink to each softball practice. At her first practice, she drank 12 bottles of sports drinks. This means she drank a total of:

12 bottles x 0.75 quarts/bottle = 9 quarts

At her second practice, she drank the entire bottle, which is another 0.75 quarts.

Therefore, the total amount of sports drinks Eden consumed during both practices are:

9 quarts + 0.75 quarts = 9.75 quarts

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Answer:

x = 55°

Step-by-step explanation:

Inscribed Angle Theorem says that the angle inside the circle is equal to 0.5 times the measurement of the arc it span. For example, the 70° spans 140° of the arc.

Create two equations to solve for x and r.

x = 0.5r

2r + 2(70) = 360

2r + 140 = 360

2r = 220

r = 110

x = 0.5 * 110

x = 55°

Since, distribution is skewed right even after we remove the outlier (600) from the data set. The median is a good measure regardless of whether the outlier is included.

We have a data set represents the dollar amounts of donations collected at the entrance to a free museum during one hour. The data present in above figure. Now, first we determine the median. As we know, median of a data set is equals to the middle value of data set, so sometimes it is called middle value. It is the value which separates upper and lower halves of data sample or population etc. First we ordered the data values in ascending order. In case odd number of data values, median = middle value and in case of even number of terms, median = mean of two middle terms. Now, here data values in ascending order are 1,5,5,5,5,5,10,10,10,15,600. That is total data values = 11 ( odd) , so median

= 6th value = 5.

So, regardless of whether the outlier is included median is good measure in this situation.

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The appropriate test of hypothesis in this case would be a two-tailed test.

A two-tailed test is used when the hypothesis being tested is that there is a difference between the groups being compared, but the direction of the difference is not specified. In this case, the hypothesis being tested is whether sex education classes and free clinics that offer counseling for teenagers reduce the number of pregnancies among teenagers.

The hypothesis does not specify the direction of the difference, only that there is a difference. Therefore, a two-tailed test would be appropriate to determine whether there is a significant difference in the number of pregnancies among teenagers who have access to sex education classes and free clinics compared to those who do not have access.

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The sample size required to provide a 95% confidence interval with a margin of error of 3, assuming a population standard deviation of 50, is approximately 106.

To calculate the sample size needed for a confidence interval, we use the formula:

n = (Z² × σ²) / E²

where:

n = sample size

Z = Z-value (corresponding to the desired confidence level)

σ = population standard deviation

E = margin of error

For a 95% confidence interval, the Z-value is 1.96. Plugging in the values, we get:

n = (1.96² × 50²) / 3²

n = 3841 / 9

n = 426.78

Since we need a whole number, we round up to the nearest integer, giving us a sample size of 107.

Therefore, the sample size required to provide a 95% confidence interval with a margin of error of 3, assuming a population standard deviation of 50, is approximately 106.

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Answer:

x+48=180-2x (straight lines sum is 180 and corresponding angles)

x+2x=180-48

3x=132

x=44°

we can represent this as a trinomial with a perfect square:(y-10)²-100

PRECISE SQUARE TERMINAL: WHAT IS IT?

A quadratic expression that can be factored into a binomial squared is a perfect square trinomial. When it is factored, it follows a pattern where the first and last terms are monomial perfect squares and the middle term is the product of the first two. A given trinomial is not a perfect square trinomial if the pattern does not fit for it1.

For instance, the trinomial x2+6x+9 is a perfect square trinomial since it factors into (x+3)². The square of x is represented by the first term, x², and the square of 3 by the last term, 9. Its result 2(x)(3) is twice as long as the middle term 6x.

Y has a coefficient of -20. A fraction of -20 is -10.

-10 is squared to give 100.

As a result, we can add and take 100 to finish the square.

It is necessary to add and subtract the square of the y-half coefficient's in order to finish the square for y²–20y.

y²-20y+100-100

=(y-10)²-100

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The study design described is a cross-sectional study.

The study design in this scenario is a cross-sectional study. The study aims to examine the relationship between family status and school performance. The sample is selected by randomly choosing every 5th child from the list of children at the ABC elementary school. Data is collected from the school record for each child, and family status is operationalized by whether the child lives in a single-parent household, dual-parent household, or has a non-parent guardian in the household. The school performance is measured by administering a student math test and a student reading test to the child in April, with each test including 20 questions and each correct answer being worth five points.

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Answer:

The interval estimate is:

41.5 - 2.45 ≤ true average time ≤ 41.5 + 2.45

or

38.05 ≤ true average time ≤ 44.95

Inequality notation:

[38.05, 44.95]

The interval estimate that will estimate the true average time that people spend brushing their teeth is [39.05, 43.95] seconds.

Given that the mean time spent brushing teeth from the sample is 41.5 seconds, and the margin of error is 2.45 seconds, we can construct a confidence interval estimate using the formula:

(sample mean) ± (margin of error)

Substituting the given values, we get:

41.5 ± 2.45

Simplifying, we get:

[39.05, 43.95]

Therefore, we can be 95% confident that the true average time that all people spend brushing their teeth falls within this interval.

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The null hypothesis for the alternative hypothesis "the percentage of internet users who use the internet for shopping is greater than 0.40" can be expressed as:
H0: p ≤ 0.40

Here, H0 represents the null hypothesis and p represents the proportion of internet users who use the internet for shopping.

The null hypothesis states that the percentage of internet users who use the internet for shopping is less than or equal to 40%.

In hypothesis testing, we collect data and analyze it to determine whether we can reject the null hypothesis or not.

If the data provides strong evidence against the null hypothesis, we reject it and accept an alternative hypothesis that suggests there is a significant relationship or difference.

However, if the data is not strong enough to reject the null hypothesis, we fail to reject it, but we cannot conclude that the null hypothesis is true.

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The equations that are quadratic functions are 6 x - 1 = 4 x + 7 x^2 and x^2 + 5 x - 7 = 0

The quadratic equation is an equation of the form:

ax^2 + bx + c = 0

where a, b, and c are constants, and x is the variable.

Out of the given equations, the following are quadratic equations:

6 x - 1 = 4 x + 7 x^2

x^2 + 5x - 7 = 0

The other equations are not quadratic equations:

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Answer:

no solutions

Step-by-step explanation:

y + 7 = y - 3 ( subtract 7 from both sides )

y = y - 10 ( subtract y from both sides )

0 = - 10 ← false statement

this false statement indicates the equation has no solution

Answer: 63

Step-by-step explanation:

tan(B) = 10/5

B = tan^-1 (10/5)

B = 63.43 ≈ 63